poj2501

/* * poj2501.cpp * * Created on: 2010-8-5 * Author: friendy */ #include #include #include using namespace std; //这道题目主要考虑在开始没速度到时候要输出距离0 //另外距离要用double类型 int main() { int i, sp, time, timeNow, len; double dis = 0; char s[100]; //读入第一次的速度 while (gets(s) != NULL) { if (strlen(s) > 8) break; else printf("%s 0.00 km/n", s); } //第一次到时间和速度 time = ((s[0] - '0') * 10 + s[1] - '0') * 3600 + ((s[3] - '0') * 10 + s[4] - '0') * 60 + ((s[6] - '0') * 10 + s[7] - '0'); sp = 0; i = 8; len = strlen(s); while (i < len) { if (s[i] >= '0' && s[i] <= '9') { sp = sp * 10 + (s[i] - 48); } i++; } //处理每一个过程 while (gets(s) != NULL) { len = strlen(s); timeNow = ((s[0] - '0') * 10 + s[1] - '0') * 3600 + ((s[3] - '0') * 10 + s[4] - '0') * 60 + ((s[6] - '0') * 10 + s[7] - '0'); dis += (double)((timeNow - time) * sp) / 3600.0; if (len == 8) { printf("%s %.2f km/n", s, dis); } else { i = 8; sp = 0; while (i < len) { if (s[i] >= '0' && s[i] <= '9') { sp = sp * 10 + (s[i] - 48); } i++; } } time = timeNow; } return 0; }